Question

这是早期question此处

的扩展

我有两张桌子，昨天和今天，每张桌子都有两列

StockNumber, StockLevel

我如何找到

只有StockLevel中存在差异的StockNumbers
退回受影响的StockNumber＆amp; StockLevel值

例如：

今日表：

StockNumber, StockLevel
ABC, 10
CBA,5
123,5

昨天表：

StockNumber, StockLevel
ABC, 9
CBA,5
123,10

返回以下wi

StockNumber, StockLevel
ABC, 10
123, 5

现在注意到StockNumber“ABC”今天返回值10，StockNumber“123”也返回今天的值5，StockNumber“CBA”没有变化，也没有返回。

非常感谢，

马特

Answer 1

这将适用于两天都存在的股票;

SELECT t.StockNumber, t.StockLevel
FROM today t
JOIN yesterday y
  ON t.StockNumber=y.StockNumber
 AND t.StockLevel <> y.StockLevel

如果您（如您之前的问题中所述）也希望包含今天的新股票;

SELECT t.StockNumber, t.StockLevel
FROM today t
LEFT JOIN yesterday y
  ON t.StockNumber = y.StockNumber
WHERE y.StockNumber IS NULL 
   OR t.StockLevel <> y.StockLevel

An SQLfiddle to test both with

Answer 2

您可以加入StockNumber上的表格并选择那些行WHERE NOT A.StockLevel = B.StockLevel。

Answer 3

这有点棘手，因为您可能在两个表中有不同的项目。我建议使用union all和group by：

select stockNumer,
       max(case when which = 'yesterday' then stocklevel end) as yesterdaylevel,
       max(case when which = 'today' then stocklevel end) as todaylevel
from ((select StockNumber, StockLevel, 'yesterday' as which
       from yesterday
      ) union all
      (select StockNumber, StockLevel, 'today' as which
       from today
      )
     ) t
group by stockNumber
having min(StockLevel) <> max(stockLevel) and count(*) = 2

Answer 4

select T.StockNumber, T.StockLevel 
from Today as T 
left join Yesterday as Y 
on T.StockNumber = Y.StockNumber 
WHERE T.StockLevel <> Y.StockLevel

在此处播放结果：http://sqlfiddle.com/#!2/367b0/3/0

Answer 5

以下查询将返回与昨天不同（从StockLevel开始）或昨天不存在的股票（这是LEFT和IS NULL的用途）：< / p>

SELECT Today.*
FROM Today LEFT JOIN Yesterday
    ON Today.StockNumber = Yesterday.StockNumber
WHERE
    Today.StockLevel <> Yesterday.StockLevel
    OR Yesterday.StockLevel IS NULL

Mysql仅返回两个表之间存在差异的记录

5 个答案: